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Session I.2.8

Part I Review of Fundamentals

Module 2 Basic Physics and MathematicsUsed in Radiation Protection

Session 8 Decay Chains and Equilibrium

IAEA Post Graduate Educational CourseRadiation Protection and Safety of Radiation Sources

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Introduction

Radioactive serial decay and equilibrium will be discussed

Students will: learn the differences between secular and

transient equilibrium identify when no equilibrium is possible understand how series decay works calculate ingrowth of a decay product from a

radioactive parent

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Content

Secular equilibrium

Transient equilibrium

Case of no equilibrium

Radioactive decay series

Ingrowth of decay product from a parent radionuclide

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Overview

Radioactive decay chains (parent and single decay product) and equilibrium situations will be discussed

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Types of Radioactive Equilibrium

Secular Half-life of parent much greater (> 100 times) than that of decay product

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Types of Radioactive Equilibrium

Transient Half-life of parent only a little greater than that of decay product

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90Sr 90Y 90Zr

Sample Radioactive Series Decay

where 90Sr is the parent (half-life = 28 years)

and 90Y is the decay product (half-life = 64 hours)

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Differential Equation forRadioactive Series Decay

= Sr NSr - Y NY dNY

dt

Parent and Single Decay Product

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Parent and Single Decay Product

Differential Equation forRadioactive Series Decay

NY(t) = (e- t - e- t)Sr YSrNSr

Y - Sr

o

Recall that Sr NoSr = Ao

Sr which equals the initial activity of 90Sr at time t = 0

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General Equation forRadioactive Series Decay

YNY(t) = (e- t - e- t)Sr Y

Y - Sr

Y SrNSro

Activity of 90Sr at time t = 0

Activity of 90Y at time t or AY(t)

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Buildup of a Decay Product underSecular Equilibrium Conditions

Secular Equilibrium

AY(t) = (1 - e- t)YASr

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Secular Equilibrium

SrNSr = YNY

ASr = AY

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Decay of226Ra to 222Rn

Secular Equilibrium

ARn (t) = Ao (1 - e- t ) RnRa

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226Ra (half-life 1600 years) decays to 222Rn (half-life 3.8 days). If initially there is 4000 kBq of 226Ra in a sample and no 222Rn, calculate how much 222Rn is produced:

a. after 7 half-lives of 222Rnb. at equilibrium

Sample Problem 1

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The number of atoms of 222Rn at time t is given by:

Solution to Sample Problem

= Ra NRa - Rn NRn dNRn

dt

Solving:

NRn(t) = (1 - e- t)RnRaNRa

Rn

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Multiplying both sides of the equation by Rn:

ARn(t) = ARa (1 - e- t)Rn

Solution to Sample Problem

= 4000 x (0.992) = 3968 kBq of 222Rn

Let t = 7 TRn

Rnt = (0.693/TRn) x 7 TRn = 0.693 x 7 = 4.85

e-4.85 = 0.00784

ARn (7 half-lives) = 4000 kBq x (1 - 0.00784 )

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Solution to Sample Problem

4000 kBq + 4000 kBq = 8000 kBq

RnNRn = RaNRa or ARn = ARa = 4000 kBq

Note that the total activity in this sample is:

RnNRn + RaNRa or ARn + ARa =

Now, at secular equilibrium:

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Transient Equilibrium

DND = D - P

D P NP

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Transient Equilibrium

AD = D - P

AP D

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Time for Decay Productto Reach Maximum Activity

Transient Equilibrium

tmD = D - P

lnD

P

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Example ofTransient Equilibrium

132Te Decays to 132I

Transient Equilibrium

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The principle of transient equilibrium is illustrated by the Molybdenum-Technetium radioisotope generator used in nuclear medicine applications.

Given that the generator initially contains 4000 MBq of 99Mo (half-life 66 hours) and no 99mTc (half-life 6 hours) calculate the:

a. time required for 99mTc to reach its maximum activityb. activity of 99Mo at this time, andc. activity of 99mTc at this time

Sample Problem

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Note that only 86% of the 99Mo transformations produce 99mTc. The remaining 14% bypass the isomeric state and directly produce 99Tc

Sample Problem

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Tc = 0.693/(6 hr) = 0.12 hr-1

Mo = 0.693/(66 hr) = 0.011 hr-1

Solution to Sample Problem

tmTc = Tc - Mo

lnTc

Mo

tmTc = 0.12 – 0.011

ln0.12

0.011= 21.9 hrs

a)

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(b) The activity of 99Mo is given by

A(t) = Ao e-t = 4000 x e(-0.011/hr x 21.9 hr)

= 4000 x (0.79) = 3160 MBq

Solution to Sample Problem

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c) The activity of 99mTc at t = 21.9 hrs is given by:

Solution to Sample Problem

ATc(t) = (e-(0.011)(21.9) - e-(0.12)(21.9))(0.12 – 0.011)

(0.12)(4000 MBq)(0.86)

= (3787) (0.785 - 0.071) = 2704 MBq of 99mTc

ATc(t) = (e- t - e- t )Mo TcTc - Mo

TcAMo(see slide 10)

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Solution to Sample Problem

The maximum activity of 99mTc is achieved at 21.9 hours which is nearly 1 day.

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Types of Radioactive Equilibrium

No Equilibrium Half-life of parent less than that of decay product

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No Equilibrium

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Summary

Secular equilibrium was defined Transient equilibrium was defined Case of no equilibrium was defined Series decay equations were developed Decay examples were discussed Problems in secular and transient

equilibrium were solved

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Where to Get More Information

Cember, H., Johnson, T. E., Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2008)

Martin, A., Harbison, S. A., Beach, K., Cole, P., An Introduction to Radiation Protection, 6th Edition, Hodder Arnold, London (2012)

Jelley, N. A., Fundamentals of Nuclear Physics, Cambridge University Press, Cambridge (1990)

Firestone, R.B., Baglin, C.M., Frank-Chu, S.Y., Eds., Table of Isotopes (8th Edition, 1999 update), Wiley, New York (1999)